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Each point in a circle is colored in one of 3 colors (blue, White, or red). Prove that one can find points that are vertices of an isosceles triangle, and either 3 points are all colored with the same color or three points are colored with 3 different colors.

I have no clue were to start, I think i have to apply pigeon hold principle to this questions along with Sperners lemma but i have no idea where to begin.

Any help is appreciated.

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    $\begingroup$ Since there are an infinite number of points, we can consider the circle to be an interval on the real numbers. For simplicity, say $[0,1]$. Then, the equivalent to an isosceles triangle is three points, $x_1, x_2, x_3$, such that $|x_1 - x_2| = |x_3 - x_2|$. $\endgroup$ – Cole Hansen Oct 12 '14 at 20:36
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    $\begingroup$ Got that, but how do I apply pigeon hold principle and the lemma. $\endgroup$ – OLE Oct 12 '14 at 20:39
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Assume otherwise and consider a regular pentagon $ABCDE$ of the circle. By pigeonhole we find a monochromatic edge or diagonal. Note that any triangle with vertices $\in\{A,B,C,D,E\}$ is isosceles (there are only two lengths: edge and diagonal). If any of the colours occurs three times, this gives us a monochromatic isosceles triangle. Hence we may assume that each colour occurs at most twice. Then each colour occurs at least once. This gives us a trichromatic isosceles triangle.

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